Time Response of Control Systems

First Order System

Step Input FO

First Order Systems have the general transfer function of:

$$\frac{X(s)}{u(s)}=\frac{K}{s+\frac{1}{\tau }}$$

Reference: TF

And the Laplace for our step input is:

$$U(s)=\frac{a}{s}$$

Reference: Laplace of Step Input

Which means:

$$\begin{array}{rl}\frac{X(s)}{\frac{a}{s}}& =\frac{K}{s+\frac{1}{\tau }}\\ X(s)& =\frac{K}{s+\frac{1}{\tau }}\ast \frac{a}{s}\\ & =\frac{Ka}{s(s+\frac{1}{\tau })}\\ & \\ \text{Employing }& \text{Partial Fractions:}\\ X(s)& =K\tau a(\frac{1}{s}-\frac{1}{s+\frac{1}{\tau }})\end{array}$$

Then we can take the inverse Laplace Transform to express this in the time domain as:

$$x(t)=K\tau a(1-e^{-\frac{1}{\tau }})$$

We can also deduce that as $x\to \infty$ the value of this becomes:

$$x(\infty )=K\tau a$$

Second Order

Step Input SO

$$x(\infty )=\frac{Ka}{{\omega }_n^2}$$

See: Note 1.2.2, Page 8

This can be calculated using the Final Value Theorem: $x(\infty )={\lim }_{t\to \infty }x(t)={\lim }_{s\to 0}sX(s)$ Which means that:

$$\begin{array}{rl}\frac{X(s)}{U(s)}& =\frac{K}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2},U(s)=\frac{a}{s}\\ & \\ \therefore X(s)& =\frac{Ka}{s(s^2+2\xi {\omega }_{n}s+{\omega }_n^2)}\\ & \\ \therefore x(\infty )& =\underset{s\to 0}{\lim }\frac{sKa}{s(s^2+2\xi {\omega }_{n}s+{\omega }_n^2)}\\ & =\underset{s\to 0}{\lim }\frac{Ka}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}\\ & =\frac{Ka}{{\omega }_n^2}\end{array}$$